D3-2. Magnetic Force between Two Parallel Conductors

1. Force Between Two Parallel Wires

The Formula

$$\frac{F}{L} = \mu_0 \frac{I_1 I_2}{2\pi r}$$
  • Calculates the magnetic force per unit length between two long, straight, parallel wires.
  • Derived using Newton's Third Law: The forces on each wire are always equal in magnitude and opposite in direction.

Notation Key & Units:

  • $F/L$ = Force per unit length ($\text{N m}^{-1}$)
  • $\mu_0$ = Permeability of free space ($4\pi \times 10^{-7}\text{ T m A}^{-1}$)
  • $I_1, I_2$ = Currents in wire 1 and wire 2 ($\text{A}$)
  • $r$ = Separation distance between wires ($\text{m}$)

Major Exam Trap!

The magnetic force between parallel wires behaves the opposite of electric charges!

  • Currents in the same direction attract.
  • Currents in opposite directions repel.
WIRES WITH CURRENTS IN OPPOSITE DIRECTIONS WILL REPEL WIRES WITH CURRENTS IN THE SAME DIRECTION WILL ATTRACT

2. Examples

Example 1

Problem The diagram shows two current-carrying wires, $P$ and $Q$, that both lie in the plane of the paper. The arrows show the conventional current direction in the wires.

S A. B. C. D. out of page into page

The electromagnetic force on $Q$ lies in the same plane as the wires. What is the direction of the electromagnetic force acting on $Q$?

Solution:

Because the currents are travelling in opposite directions, the wires will repel each other. Therefore, the force on wire $Q$ will point away from wire $P$.

Example 2

Problem The diagram shows two parallel wires $X$ and $Y$ that carry equal currents into the page.

45° 45° X Y Q

Point $Q$ is equidistant from the two wires. The magnetic field at $Q$ due to wire $X$ alone is $15\text{ mT}$.
(i) On the diagram draw an arrow to show the direction of the magnetic field at $Q$ due to wire $X$ alone.
(ii) Determine the magnitude and direction of the resultant magnetic field at $Q$.

Solution:

(i) Using the Right-Hand Grip Rule, the field around $X$ (current into page) is clockwise. At point $Q$, the tangent points perpendicularly downwards and to the right.
(ii) Wire $Y$ also has current into the page. By symmetry, its field at $Q$ has the same magnitude ($15\text{ mT}$) but points downwards and to the left. The horizontal components cancel perfectly. The vertical components add together to create a resultant field pointing straight down. Magnitude can be calculated using trigonometry based on the specific angles provided in the diagram.

Example 3

Problem Two long, straight, parallel wires are separated by a distance of $d = 2.0\text{ cm}$. Wire 1 carries a current of $I_1 = 5.0\text{ A}$ and Wire 2 carries a current of $I_2 = 3.0\text{ A}$. Initially, both currents flow in the same direction.

Wire 1 (5.0 A) Wire 2 (3.0 A) d = 2.0 cm

(a) Calculate the magnitude of the magnetic force per unit length between the wires. Is this force attractive or repulsive?
(b) If the current in Wire 2 is reversed so that the currents now flow in opposite directions, what is the new magnitude and direction of the force?

Solution:

(a) The formula for the magnetic force per unit length ($\dfrac{F}{L}$) between two parallel current-carrying wires is given by:

$$ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} $$

Where the permeability of free space is $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$. Convert the distance to meters ($d = 0.020\text{ m}$) and substitute the values:

$$ \frac{F}{L} = \frac{(4\pi \times 10^{-7}\text{ T}\cdot\text{m/A})(5.0\text{ A})(3.0\text{ A})}{2\pi (0.020\text{ m})} $$ $$ \frac{F}{L} = \frac{2 \times 10^{-7} \times 15}{0.020}\text{ N/m} = 1.5 \times 10^{-4}\text{ N/m} $$

Because the currents are flowing in the same direction (like currents), the resulting magnetic force between the wires is attractive.

(b) Reversing the direction of the current in Wire 2 does not change the magnitude of the currents or the distance between the wires. Therefore, the magnitude of the force per unit length remains exactly the same:

$$ \frac{F}{L} = 1.5 \times 10^{-4}\text{ N/m} $$

However, because the currents are now flowing in opposite directions (unlike currents), the magnetic field interactions change, and the force between the wires becomes repulsive.